Answer:
4+2 then 3+7 then what is 10x6=60 so 60 is the area
Step-by-step explanation:
Find out the determinant
we have
2x-y=4
3x+y=1
so
D=(2)(1)-(-1)(3)
D=2+3
D=5
therefore
<h2>the answer is the second option</h2><h2>Verify</h2>
solution (1,-2)
equation 1 -------> 2(1)-(-2)=4 -------> 4=4 is true -----> is ok
equation 2 ------> 3(1)-2=1 -----> 1=1 ----> is true -----> is ok
This is just simple. For example you have a plane of the form x=a, then you just substitute x with a, and you'll get an equation with y and z only, hence you have a 2-d trace of the intersection. It is just similar for y=b and z=c.
(1) At z=1.5, 2x^2 + 5y^2 + 1.5^2 = 4
2x^2 + 5y^2 = 1.75
Now you have an ellipse in the z=1.5 plane as your trace.
(2) At x=1, 2(1)^2 + 5y^2 + z^2 = 4
5y^2 + z^2 = 2
Now you have an ellipse in the x=1 plane as your trace.
(3) At z=0, 2x^2 + 5y^2 + (0)^2 = 4
2x^2 + 5y^2 = 4
Now you have an ellipse in the z=0 plane as your trace.
(4) At y=0, 2x^2 + 5(0)^2 + z^2 = 4
2x^2 + z^2 = 4
Now you have an ellipse in the y=0 plane as your trace.
Answer:
1) x ≤ 2 or x ≥ 5
2) -6 < x < 2
Step-by-step explanation:
1) We have x^2 - 7x + 10, so let's factor this as if this were a regular equation:
x^2 - 7x + 10 = (x - 2)(x - 5)
So, we now have (x - 2)(x - 5) ≥ 0
Let's imagine this as a graph (see attachment). Notice that the only place that is above the number line is considered greater than 0, and that's when x ≤ 2 or x ≥ 5 (the shaded region).
2) Again, we have x^2 + 4x - 12, so factor this as if this were a regular equation:
x^2 + 4x - 12 = (x + 6)(x - 2)
So now we have (x + 6)(x - 2) < 0
Now imagine this as a graph again (see second attachment). Notice that the only place that is below 0 (< 0) is when -6 < x < 2 (the shaded region).
Hope this helps!
